# Solvintg Perl Weekly Challenge 087

Here comes Perl Weekly Challenge 087.

## TASK #1 › Longest Consecutive Sequence

``````You are given an unsorted array of integers @N.

Write a script to find the longest consecutive sequence. Print 0 if none sequence
found.

Example 1:

Input: @N = (100, 4, 50, 3, 2)
Output: (2, 3, 4)

Example 2:

Input: @N = (20, 30, 10, 40, 50)
Output: 0

Example 3:

Input: @N = (20, 19, 9, 11, 10)
Output: (9, 10, 11)
``````

It looks like the in put @N contains a bunch of unordered integers while the output should be a range of integers that is entriely contained in the input @N. If there are multiple ranges of such, pick the longest one.

If @N is already sorted by ascendeing order, we could derive the 'current range' by iterating over the array @N and check whether the current number are difference by one with previous number, as well as keep tracking if the current range is longer than the longest range we know.

Here's a Raku version, with time complexity being O(n log(n)), space complexit being O(1):

``````sub longest-consecutive-sequence(@N) {
my @m = @N.sort({ \$^a <=> \$^b });

my \$seq_from = @m;
my \$seq_until = @m;
my \$longest_seq_from = @m;
my \$longest_seq_until = @m;
my \$longest_seq_length = 0;

for 1..@m.end -> \$i {
my \$n = @m[\$i];
if \$n - @m[\$i-1] == 0|1 {
\$seq_until = \$n;
my \$len = \$seq_until - \$seq_from;
if \$longest_seq_length < \$len {
\$longest_seq_from   = \$seq_from;
\$longest_seq_until  = \$seq_until;
\$longest_seq_length = \$len;
}
} else {
\$seq_from  = \$n;
\$seq_until = \$n;
}
}

return \$longest_seq_length == 0 ?? Nil !! [\$longest_seq_from...\$longest_seq_until];
}
``````

## TASK #2 › Largest Rectangle

``````You are given matrix m x n with 0 and 1.

Write a script to find the largest rectangle containing only 1. Print 0 if none found.

Example 1:

Input:
[ 0 0 0 1 0 0 ]
[ 1 1 1 0 0 0 ]
[ 0 0 1 0 0 1 ]
[ 1 1 1 1 1 0 ]
[ 1 1 1 1 1 0 ]

Output:
[ 1 1 1 1 1 ]
[ 1 1 1 1 1 ]

Example 2:

Input:
[ 1 0 1 0 1 0 ]
[ 0 1 0 1 0 1 ]
[ 1 0 1 0 1 0 ]
[ 0 1 0 1 0 1 ]

Output: 0

Example 3:

Input:
[ 0 0 0 1 1 1 ]
[ 1 1 1 1 1 1 ]
[ 0 0 1 0 0 1 ]
[ 0 0 1 1 1 1 ]
[ 0 0 1 1 1 1 ]

Output:
[ 1 1 1 1 ]
[ 1 1 1 1 ]
``````

This looks like we should be finding a largest rectangle of 1s from a 01-matrix. A naive solution would be to enumerating all sub-matrices, removing the ones not containing all 1s, and find the largest. What defines the largest matrix ? There are no such definition from the body of the task. Let's just use the area of the matrix, which is the width multiplied by height.

A sub-matrix could be defined by two coordinates, one at the top-left corner, the other at the bottom-right. It appears to me that sub-matrices with dimensions beinge 1x1 does not count, in other words, the coordinates of two corners cannot be identical. Therefore, to enumerate all sub-matrices would be the same as taking the combination of 2 coordinates, then removing the ones that violates the (top-left, bottom-right) relationship,

Conventionally, we contain a matrix in a two dimentional array `@matrix` with the first dimension being rows, and the second dimension beingg columns. Which means the hight `\$h` would be the number of elements of the first dimension, and the width `\$w` would be the number of elements of the second.

``````\$h = @matrix.elems;
\$w = @matrix.elems;
``````

To take all coordinates of a such matrix would be to derive the cross-product of two integer sets of `[0 .. ^\$h]` and `[0 .. ^\$w]`:

``````([^\$h] X [^\$w])
``````

... then we take the combination of 2, removing the ones that violate the (top-left, bottom-right) relationship>..

``````.combinations(2)
.grep(-> (\$p, \$q) { \$p <= \$q and \$p <= \$q })
``````

... now we have a set of bi-coordinates in the form of (top-left, bottom-right). We transform that to the actual sub-matrix:

``````.map(
-> (\$p, \$q) {
(\$p .. \$q).map(
-> \$y {
@matrix[\$y][ \$p .. \$q ]
})
})
``````

... removing the ones that do not contain only 1s ...

``````.grep(
-> @m {
([^@m] X [^@m]).map(-> (\$y,\$x) { @m[\$y][\$x] == 1 }).all
}
)
``````

... and take the one with maximum value of width multiplied by height...

``````.max(-> @m { @m.elems * @m.elems });
``````

... and there we have a solution.

Here's the subroutine with all fragment combined.

``````sub largest-rectangle(@matrix) {

([^@matrix] X [^@matrix])
.combinations(2)
.grep(
-> (\$p, \$q) {
\$p <= \$q and \$p <= \$q
})

.map(
-> (\$p, \$q) {
(\$p .. \$q).map(
-> \$y {
@matrix[\$y][ \$p .. \$q ]
})
})

.grep(
-> @m {
([^@m] X [^@m]).map(-> (\$y,\$x) { @m[\$y][\$x] == 1 }).all
})

.max(
-> @m {
@m * @m
})
}
``````

Regarding the part to build the sub-matrix, out of intuition I utilize the syntax of array slicing and wrote this:

``````    @matrix[ \$p .. \$q ][ \$p .. \$q ]
``````

It failed with tons of runtime errors. I wonder if there are syntax shortcuts for slicing a multi-dimensional array as a multi-dimensional array.